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Q.Identify the oxidizing and reducing agent in the reaction: N2H4(l) + 2H2O2(l) -> N2(g) + 4H2O(l)

Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2026Subjective· 2mImportance★★★★★
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In N2H4(l) + 2H2O2(l) → N2(g) + 4H2O(l): N2H4 is the reducing agent, H2O2 is the oxidizing agent.

Track oxidation numbers of the atoms that change:

Nitrogen: In N2H4, each N is at oxidation number -2 (each H contributes +1, so with 4 H's total +4, and the molecule is neutral, so 2 N's total -4, i.e. each N = -2). In the product N2, elemental nitrogen has oxidation number 0. So N goes from -2 to 0: an increase in oxidation number, i.e., nitrogen is oxidised. The species containing it, N2H4, is therefore the reducing agent (it causes reduction elsewhere by itself being oxidised).

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